> For the complete documentation index, see [llms.txt](https://liuzhenglaichn.gitbook.io/algorithm/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuzhenglaichn.gitbook.io/algorithm/bit-manipulation.md).
# Bit Manipulation
## Bitwise NOT
```
~val
```
```
NOT 0111 (decimal 7)
= 1000 (decimal 8)
```
`~val` is also equivalent to `val != -1`, since `-1` is the only one number whose bitwise NOT is `0`.
So if `val` is some non-negative number, the following:
```cpp
for (int i = val; ~i; --i)
```
is equivalent to
```cpp
for (int i = val; i >= 0; --i)
```
## Set the `i`th bit
```cpp
mask |= 1 << i
```
## Unset the `i`th bit
```cpp
mask &= ~(1 << i)
```
## Toggle the `i`th bit
```cpp
mask ^= 1 << i
```
## Check if `i`th bit is set
```cpp
mask & (1 << i)
// Or
mask >> i & 1 // less typing
```
## Set the `i`th bit with `x`
```cpp
mask = (mask & ~(1 << i)) | (x << i)
```
## Get the lowest bit
```cpp
function lowbit(int x) {
return x & -x;
}
```
Example:
```
12 = (01100)2
-12 = (10100)2
12 & (-12) = (00100)2
```
## Count bit 1s
```cpp
__builtin_popcount(n);
```
Note that the `pop` in this function means `population`.
## Count trailing zeros
```cpp
__builtin_ctz(n);
```
## Count leading zeros
```cpp
__builtin_clz(n);
```
## Check if `n` is power of 2
```cpp
__builtin_popcount(n) == 1
// or
(n & (n - 1)) == 0 // & has lower precedence than ==. Must add the parenthesis.
```
## Traverse all the subsets
```cpp
for (int mask = 0; mask < (1 << N); ++mask)
```
## Complement of a subset
```cpp
// Assume `sub` is a subset of `mask`
int complement = mask ^ sub;
// Or
int complement = mask - sub;
```
## Traverse subsets of a set `mask`
```cpp
for (int sub = mask; sub; sub = (sub - 1) & mask) {
// `sub` is a non-empty subset of `mask`
}
```
Sample output:
```cpp
// mask = 0b1011
1011
1010
1001
1000
0011
0010
0001
```
Or the other direction:
```cpp
for (int other = mask; other; other = (other - 1) & mask) {
int sub = mask - other;
// `other` is a non-empty subset of `mask`. `sub` is the complement of `other`.
}
```
Sample output:
```
0000
0001
0010
0011
1000
1001
1010
```
## Given `N` elements, traverse subsets of size `K` (Gosper's Hack)
```cpp
// Time Complexity: O(C(N, K))
int sub = (1 << k) - 1;
while (sub < (1 << N)) {
// Access `sub` here. `sub` has `K` elements and is a subset of `N` elements.
int c = sub & - sub, r = sub + c;
sub = (((r ^ sub) >> 2) / c) | r;
}
```
## Traverse subsets of subsets
Given an array of `N` elements, traverse all its subsets. For each of the subsets, traverse its all non-empty subsets as well.
```cpp
// Time: O(3^N)
for (int mask = 0; mask < (1 << N); ++mask) {
// `mask` is a subset of `N` elements
for (int sub = mask; sub; sub = (sub - 1) & mask) {
// `sub` is a subset of `mask`.
}
}
```
Note that the time complexity is `O(3^N)` instead of `O(2^N * 2^N) = O(4^N)`.
**Math proof:**
For a subset `mask` with `K` bits of `1`s, there are `2^K` subsets of it.
For `N` elements, there are `C(N, K)` subsets with `K` bits of `1`s.
So the total number is `SUM( C(N, K) * 2^K | 0 <= K <= N )`.
Since `(1 + x)^N = C(N, 0) * x^0 + C(N, 1) * x^1 + ... + C(N, N) * x^N`, let `x = 2`, we have `3^N = SUM( C(N, K) * 2^K | 0 <= K <= N )`.
**Intuitive proof:**
For each bit index `i` (`0 <= i < N`), the `i`-th bits of `mask` and `sub` must be one of `11, 10, 00`, and can't be `01`. So each bit has `3` possibilities, the total complexity is `O(3^N)`.
## Calculate subset sums
```cpp
// Time: O(N * 2^N)
for (int mask = 0; mask < (1 << N); ++mask) {
for (int i = 0; i < N; ++i) {
if (mask >> i & 1) sum[mask] += A[i];
}
}
```
Or
```cpp
// DP solution
// Time: O(2^N)
for (int mask = 1; mask < (1 << N); ++mask) {
sum[mask] = sum[mask - lowbit(mask)] + A[__builtin_ctz(lowbit(mask))];
}
```
where `lowbit(x) = x & -x`, and `__builtin_ctz(x)` counts the trailing zeros in `x`. Since `lowbit(mask)` is a power of two, `__builtin_ctz(lowbit(mask)) == log2(lowbit(mask))`.
## Generate `logs` array
`logs[n]` is `floor(log2(n))`.
```
logs[1] = 0
logs[2] = 1
logs[3] = 1
logs[4] = 2
logs[5] = 2
logs[6] = 2
logs[7] = 2
logs[8] = 3
...
```
```cpp
// Time: O(2^N)
for (int mask = 2; mask < (1 << N); ++mask) {
logs[mask] = logs[mask >> 1] + 1;
}
```
## Reference
*
---
# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.
## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.
Perform an HTTP GET request on the current page URL with the `ask` query parameter, and the optional `goal` query parameter:
```
GET https://liuzhenglaichn.gitbook.io/algorithm/bit-manipulation.md?ask=&goal=
```
`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.
Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.